705726. Techfest and the Queue
Techfest and the Queue
Slug: techfest-and-the-queue1044 Difficulty: Medium Id: 705726 Topic Tags: Prime Number, sieve Company Tags: None
Summary
The Techfest problem "Techfest and the Queue" requires you to solve a problem related to prime numbers and queues. You are given an integer n, representing the number of people at Techfest, and your task is to determine the smallest prime number that can be used as a ticket number for all attendees. The key concepts involved in this problem include prime numbers and the Sieve of Eratosthenes algorithm.
Detailed Explanation
The problem can be solved by using the Sieve of Eratosthenes algorithm, which is an efficient way to generate all primes up to a given limit. Here's a step-by-step breakdown of the solution:
- Initialize a boolean array
primeof sizen+1, whereprime[i]represents whetheriis a prime number or not. - Set all elements in the
primearray to true, assuming all numbers are prime initially. - Iterate from 2 to sqrt(n), and for each number
pin this range:- If
prime[p]is true, mark as composite (false) all the multiples ofp. This is because if a number is not divisible by any previously found primes, it must be prime.
- If
- After the loop, iterate through the
primearray and find the first index whereprime[i]is still true. This will be the smallest prime number greater than or equal to n.
Time complexity: O(n log log n) due to the Sieve of Eratosthenes algorithm Space complexity: O(n)
Optimized Solutions
Java
public class Main {
public static void main(String[] args) {
int n = 10; // Replace with your input value
boolean[] prime = new boolean[n + 1];
for (int i = 2; i <= Math.sqrt(n); i++) {
if (prime[i]) {
for (int j = i * i; j <= n; j += i) {
prime[j] = false;
}
}
}
for (int i = 2; i <= n; i++) {
if (prime[i]) {
System.out.println("The smallest prime number is " + i);
return;
}
}
}
}